Case Study: Designing the Ascent to a 500 km Orbit

"Before you choose a rocket, you must know the number it has to beat. That number is the launch delta-v, and you build it from the orbit up."

Executive Summary

In the first case study we took a finished ascent apart. Here we do the harder thing: we build the launch requirement from scratch. Given only a target orbit — a $500\ \text{km}$ circular low Earth orbit for a small satellite — we will derive, step by step, the launch delta-v a vehicle must supply: the orbital speed at that altitude, the gravity and drag losses stacked on top, and the Earth-rotation credit a well-chosen launch site gives back. Then we will turn that delta-v into a vehicle, using the rocket equation from Chapter 3, and watch it demand staging. This is the calculation that opens every launch-vehicle trade study, and the one you will run for your own mission when you select a launcher in Chapter 30.

Skills applied

• Computing circular orbital velocity at a chosen altitude, $v = \sqrt{\mu/r}$ (§4.1).
• Assembling a launch delta-v budget from orbital speed plus gravity and drag losses (§4.3–4.5).
• Applying the Earth-rotation launch credit and comparing sites (§4.5).
• Converting a launch delta-v into a required mass ratio and proving staging is mandatory (§4.3, Chapter 3).
• Sensitivity analysis: how the budget shifts with target altitude and launch direction.

Background

The requirement

• Payload: a $300\ \text{kg}$ small satellite.
• Target orbit: circular, $500\ \text{km}$ altitude, so orbital radius $r = R_\oplus + 500 = 6{,}371 + 500 = 6{,}871\ \text{km}$.
• Deliverable: the launch delta-v the vehicle must provide, and a first sketch of the vehicle that can provide it.

We use Earth's $\mu = 3.986\times10^5\ \text{km}^3/\text{s}^2$ and $g_0 = 9.81\ \text{m/s}^2$ throughout.

Phase 1: Orbital velocity at the target altitude

Everything starts with the sideways speed the satellite must end up at. From §4.1,

$$ v_{\text{orbit}} = \sqrt{\frac{\mu}{r}} = \sqrt{\frac{3.986\times10^5}{6{,}871}} = \sqrt{58.0\ \text{km}^2/ \text{s}^2} = 7.62\ \text{km/s}. $$

Note this is slower than the $7.79\ \text{km/s}$ of a $200\ \text{km}$ orbit — higher orbits are slower (§4.1) — a fact we will return to in Phase 5, because it produces a mild surprise.

Phase 2: Stack the losses

The vehicle must supply not just the orbital speed but the delta-v the two thieves take. Using the representative figures of §4.3–4.4:

$$ \Delta v_{\text{launch}} = 7.62 + 1.5 + 0.1 = 9.22\ \text{km/s}. $$

This is the honest number a vehicle must beat if it launches from a non-rotating Earth — or, equivalently, straight north/south so Earth's spin gives no help. Now we shop for a launch site.

Phase 3: The launch-site credit

Earth rotates eastward. A due-east launch starts the vehicle already moving with the ground, and that motion counts toward orbital velocity. At the equator the surface moves east at

$$ v_{\text{spin}} = \frac{2\pi R_{\text{eq}}}{T_{\text{sidereal}}} = \frac{2\pi (6{,}378\ \text{km})}{86{,}164\ \text{s}} = 0.465\ \text{km/s}, $$

and at latitude $\phi$ the eastward speed is $v_{\text{spin}}\cos\phi$. Compare two sites:

At Cape Canaveral, $0.465\cos(28.5^\circ) = 0.465\times0.879 = 0.41\ \text{km/s}$, so the launch delta-v drops to

$$ \Delta v_{\text{launch}} = 9.22 - 0.41 = 8.81\ \text{km/s}. $$

The equatorial site does a little better ($8.76$), the high-latitude site a little worse ($8.99$). The credit is modest — a few hundred m/s — but in a game where delta-v is exponentially expensive (Chapter 3), a few hundred m/s of free velocity is worth building a spaceport near the equator to collect. (The full story of azimuth, latitude, and reachable inclinations is Chapter 30; here we take only the simplest, due-east credit.) We will design to the Cape figure, $\approx 8.8\ \text{km/s}$.

Phase 4: Turn the delta-v into a vehicle

Now Chapter 3 does the rest. Ask the naive question first: could a single stage supply $8.8\ \text{km/s}$? With a decent effective exhaust velocity $v_e = 3.0\ \text{km/s}$ (kerosene, sea-level-to-vacuum average), the required mass ratio is

$$ \frac{m_0}{m_f} = e^{\Delta v/v_e} = e^{8{,}810/3{,}000} = e^{2.94} \approx 18.8. $$

But Chapter 3 taught us that a real stage's mass ratio is capped near $1/\varepsilon \approx 12.5$ (with a structural coefficient $\varepsilon = 0.08$), because the tanks and engines have mass that cannot vanish. A mass ratio of $18.8$ is far above the ceiling — no single stage can do it. Staging is not a preference here; the arithmetic forbids the alternative.

So we split the $8.8\ \text{km/s}$ between two stages. Give the first stage $4.0\ \text{km/s}$ at $v_e = 2.9\ \text{km/s}$ (its sea-level-to-vacuum average) and the second stage the remaining $4.8\ \text{km/s}$ at $v_e = 3.4\ \text{km/s}$ (vacuum). Their mass ratios:

$$ R_1 = e^{4{,}000/2{,}900} = e^{1.38} = 3.97, \qquad R_2 = e^{4{,}800/3{,}400} = e^{1.41} = 4.10. $$

Both sit comfortably below the $12.5$ ceiling ($R_1\varepsilon = 0.32$, $R_2\varepsilon = 0.33$ — nowhere near $1$), so each stage is physically buildable. We have converted a target orbit into a feasible two-stage launch requirement:

Requirement, sized. To place a $300\ \text{kg}$ satellite in a $500\ \text{km}$ orbit from Cape Canaveral, a launch vehicle must supply $\approx 8.8\ \text{km/s}$, which a two-stage kerosene design covers with per-stage mass ratios near $4$ — well within what real structures achieve.

Phase 5: Sensitivity — the surprise, and the west penalty

Two "what ifs" sharpen the picture.

Lower the target to $200\ \text{km}$. Orbital speed rises to $7.79\ \text{km/s}$ (lower orbits are faster), so the no-credit launch delta-v is $7.79 + 1.5 + 0.1 = 9.39\ \text{km/s}$ — slightly more than the $9.22$ for $500\ \text{km}$. That is the surprise: by this simple accounting, the higher orbit looks cheaper. The catch is that our constant "$1.5\ \text{km/s}$ gravity loss" hides the extra climbing a higher orbit demands; a fuller trajectory model adds gravity loss for the additional altitude, and the $500\ \text{km}$ orbit ends up modestly more expensive after all. Flag: the additive budget captures orbital speed exactly but treats losses as a fixed lump, so it is trustworthy to a few hundred m/s, not to tens.

Launch west instead of east. Now Earth's rotation works against you: the credit becomes a penalty of $+0.41\ \text{km/s}$, and the launch delta-v jumps to $9.22 + 0.41 = 9.63\ \text{km/s}$ — nearly $0.9\ \text{km/s}$ worse than the eastward Cape launch. This is why essentially no one launches retrograde unless a specific orbit demands it: you pay twice the rotation credit, once in forfeiting it and once in fighting it.

💡 Intuition: The launch delta-v is a stack, and you build it from the orbit up: start with the sideways speed the payload must have, add the gravity and drag the climb will cost, and subtract the running start the spinning Earth hands you if you launch with its rotation. Only when that number is in hand can you shop for a rocket — because the rocket equation turns that one number into the entire size of the vehicle.

Discussion Questions

1. The equatorial and high-latitude sites differ by only $\sim 0.23\ \text{km/s}$ in rotation credit. Given how exponentially delta-v costs scale (Chapter 3), is that difference worth relocating a launch program? What non-delta-v factors also matter?
2. Phase 5 found the $200\ \text{km}$ orbit "cheaper" by the additive model but flagged it as misleading. Explain physically why climbing to $500\ \text{km}$ really does cost more, despite the lower orbital speed.
3. Why does splitting the $8.8\ \text{km/s}$ into two stages of $\sim 4\ \text{km/s}$ each keep both mass ratios far below the ceiling, whereas one stage of $8.8\ \text{km/s}$ blows past it? Tie your answer to the exponential.
4. If the customer's satellite grew from $300\ \text{kg}$ to $600\ \text{kg}$, would the launch delta-v change? Would the vehicle mass change? Explain the difference.

Your Turn: Extensions

• Option A (design). Redo Phases 1–4 for a $1{,}000\ \text{km}$ orbit ($r = 7{,}371\ \text{km}$). Compute the new orbital speed and launch delta-v, and check that a two-stage split still keeps both mass ratios under $12.5$.
• Option B (computation). Extend astrotools/rocket.py: write launch_delta_v(v_orbit, grav, drag, rot) and chain it with mass_ratio(dv, ve) to print the launch delta-v and the single-stage mass ratio for the $500\ \text{km}$ case. Add # Expected output: by hand; do not run it.
• Option C (your mission). For your mission's parking orbit, compute the orbital speed at its altitude, build the launch delta-v budget with a credit for your chosen launch site, and record the result as the top line of your MDR delta-v budget.

Key Takeaways

1. Build the launch delta-v from the orbit up: orbital speed $\sqrt{\mu/r}$ + gravity loss + drag loss − rotation credit. For a $500\ \text{km}$ orbit from the Cape, that is $\approx 8.8\ \text{km/s}$.
2. The rotation credit is real but modest (a few hundred m/s, largest at the equator, eastward) — and it becomes a penalty of the same size if you launch west.
3. The launch delta-v forces staging: at $v_e = 3\ \text{km/s}$ it demands a mass ratio near $19$, far above the $\sim 12.5$ single-stage ceiling; two stages of $\sim 4\ \text{km/s}$ each are easily feasible.
4. The additive budget is good to a few hundred m/s, not to tens — it captures orbital speed exactly but lumps the losses; trust it for a first cut, refine it with a trajectory model for the real design.